Perturbation approximation of solutions of a nonlinear inverse problem arising in olfaction...

J. Math. Biol. (2007) 55:745–765DOI 10.1007/s00285-007-0104-8 Mathematical Biology

Perturbation approximation of solutions of a nonlinearinverse problem arising in olfaction experimentation

Donald A. French · David A. Edwards

Received: 30 January 2006 / Published online: 23 June 2007© Springer-Verlag 2007

Abstract In this paper, a mathematical model of the diffusion of cAMP into olfactorycilia and the resulting electrical activity is presented. The model, which consists oftwo nonlinear differential equations, is studied using perturbation techniques. Theunknowns in the problem are the concentration of cAMP, the membrane potential, andthe quantity of most interest in this work: the distribution of CNG channels along thelength of a cilium. Experimental measurements of the total current during this diffusionprocess provide an extra boundary condition which helps determine the unknowndistribution function. A simple perturbation approximation is derived and used to solvethis inverse problem and thus obtain estimates of the spatial distribution of CNG ionchannels along the length of a cilium. A one-dimensional computer minimization anda special delay iteration are used with the perturbation formulas to obtain approximatechannel distributions in the cases of simulated and experimental data.

Keywords Olfaction · Inverse problem · Cilia · Perturbation analysis ·Computational neuroscience

Mathematics Subject Classification (2000) 92B05 · 35Q80

D. A. French (B)Department of Mathematical Sciences, University of Cincinnati,P.O. Box 210025, Cincinnati, OH 45221-0025, USAe-mail: [email protected]

D. A. EdwardsDepartment of Mathematical Sciences, University of Delaware,511 Ewing Hall, Newark, DE 19716, USAe-mail: [email protected]

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746 D. A. French, D. A. Edwards

1 Introduction

Identification of detailed features of neuronal systems is an important challenge inthe biosciences today. Cilia are long thin cylindrical structures that extend from theolfactory receptor neurons into the nasal mucus. Transduction of an odor into anelectrical signal occurs in the membranes of the cilia. The cyclic-nucleotide-gated(CNG) channels which reside in the ciliary membrane are activated by cAMP, allow adepolarizing influx of Ca2+ and Na+, and are thought to initiate the electrical signal [4].

In [2] a mathematical model for the diffusion of cAMP into an olfactory ciliumfrom a grass frog is developed and used with numerical computations to determineplausible distributions of CNG ion channels. A key conclusion that is suggested in[2] and demonstrated more strongly in [3] is that the distribution of the ion channels,ρ(x), is heavily clustered in a short segment of the cilium, roughly 1/3 of the distancefrom the base (open end) to the tip (closed end).

As is typical of inverse problems this one appears to be highly ill-conditioned;certain matrices that arise in the numerical solution process have condition numbersas high as 1013. Thus, even though the residuals and sample computations in [2] arequite accurate, it is desireable to develop solutions in a different way.

In this paper we use perturbation techniques to develop an approximate solutionto the inverse problem. Based on previous work [3], we look for ρ in the form of adensity pulse: either a delta function or a tall Gaussian with a narrow base. Thus theproblem reduces to finding the strength and position of each pulse. Using a simple one-dimensional numerical minimization and special delay iteration we develop accurateapproximations using the perturbation solution to cases with simulated and experi-mental data. Below we describe the mathematical model.

Governing equations

We begin with the dimensional equations governing our model, which we obtain from[2]. We consider diffusion and reaction of cAMP molecules in solution, the (volume)concentration of which we denote by C(x, t). cAMP diffuses in the x-direction in aone-dimensional channel of length L . Conservation of mass of cAMP is then given by

∂C

∂ t= D

∂2C

∂ x2 − α∂ S

∂ t, 0 ≤ x ≤ L , (1)

where D is the diffusion coefficient, S(x, t) is the concentration of bound cAMP, andα is a conversion factor. The equation for S is given by

S = BS ρF(C),

where BS (units of molecules per channel) is the number of binding sites that areneeded to activate the channel, and ρ(x) is the density of channels along the surfaceof the cilium. (In the inverse problem, we are trying to solve for ρ.) F(C) is thedimensionless Hill function describing the binding process, given by

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Perturbation approximation of solutions of a nonlinear inverse problem 747

F(C) = Cn

Cn + K n1/2

,

where K1/2 is the “half-maximal concentration”, so-called because it is the valueof C at which F = 1/2. From the appendix, we see that n = 1.7. Note that uponsubstitution of S into (1), we are left with a diffusion-type equation solely in C . Thus,it suffices to impose boundary and initial conditions only on C , not on S. Initially, thechannel has no cAMP, so

C(x, 0) = 0. (2)

At the exposed end of the cilium (x = 0), the concentration is the same as the valueCb in the bulk and there is no flux through the sealed end of the cilium (x = L):

C(0, t) = Cb and∂C

∂ x(L , t) = 0. (3)

The binding initiates a change in the sodium current flux J (x, t) through the channels,which is governed by the following equation:

J = gCNG PV S

BS,

where gCNG is the conductance of the channels, V (x, t) is the membrane potential,and P is the maximum open probability for the channel.

The membrane potential V is also related to the current through standard cabletheory:

1

Ra

∂2V

∂ x2 = J , 0 ≤ x ≤ L , (4)

where Ra is the (lineal) resistance density of the longitudinal current in the cilium. Theform of (4) dictates that we impose two boundary conditions upon V . At the exposedend of the cilium (x = 0), the potential is the same as the value in the bulk, which wedenote by −Vb because it is negative. Moreover, there is no change in potential at thesealed end of the cilium (x = L):

V (0, t) = −Vb and∂ V

∂ x(L , t) = 0. (5)

The quantity actually measured is I (t), the total current, which is simply the integralof J :

I =L∫

0

J d x = − 1

Ra

∂ V

∂ x(0, ·), (6)

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where we have used (4) and (5). Thus, given current data I (t) and parameters D,α, BS , K1/2, Cb, gCNG, P , Ra, and Vb we wish to find the functions C , V , and ρ.We specify that ρ consists of a single narrow pulse; thus the real unknowns are thepulse position x0 and its strength, which is equivalent to the number of channels. Thisis an inverse problem since the coefficient ρ is unknown.

In Sect. 2 we scale (1)–(6). In Sect. 3 we solve the resulting system using pertur-bation techniques. Finally, in Sect. 4 we use the perturbation solution along with bothsimulated and real current data to determine an approximation for ρ. This last sectioninvolves some basic numerical computations.

2 Scaling

In this section we nondimensionalize the governing Eqs. (1)–(6); we will introducescalings to simplify our problem and illustrate key physical ideas. We begin by scalingthose variables for which characteristic values are self-evident:

C(x, t) = C(x, t)

Cb, V (x, t) = V (x, t)

Vb, and x = x

L.

For the Hill function we rescale to obtain

F(C) = Cn

Cn + ε=

(1 + ε

Cn

)−1with ε =

(K1/2

Cb

)n

.

We choose the letter ε because from the Appendix we see that ε � 1. We note thatF(C) = O(1) for all C . We then scale the boundary and initial conditions:

C(x, 0) = 0, C(0, t) = 1,∂C

∂x(1, t) = 0, (7)

and

V (0, t) = −1,∂V

∂x(1, t) = 0. (8)

Now we can determine the proper scale for the current. Letting

I (t) = I (t)

Ic

(where the subscript “c” refers to “characteristic value”), we have

Ic I = − Vb

Ra L

∂V

∂x(0, t) or I = −∂V

∂x(0, t) where Ic = Vb

Ra L. (9)

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To scale ρ, we use its average value as the characteristic value ρc. Thus we have

ρ(x) = ρ(x)

ρcwhere ρc = 1

L

L∫

0

ρ(x) dx .

Using the definitions of J and S in (4) and then rescaling we obtain

∂2V

∂x2 = [bρF(C)]V, with b = Ra L2gCNG Pρc. (10)

Note that we have now eliminated J from our equations completely. It is shown in theAppendix that b = O(1). Though not necessary because of the form of (10), it willbe convenient later to calculate V (x, 0). Substituting t = 0 into (10), we obtain

∂2V

∂x2 (x, 0) = [bρF(C(x, 0))] V (x, 0) = 0,

where we have used (7). Then solving the above subject to (8), we have

V (x, 0) = −1. (11)

Lastly, we may use our results to create a time scale. We substitute our formula for Sinto (1) and using t = t/tc yields:

∂C

∂t= Dtc

L2

∂2C

∂x2 − αρc

CbBSρ

∂ (F(C))

∂t.

Choosing tc = L2/D, we find

∂C

∂t= ∂2C

∂x2 − aρ∂ (F(C))

∂twith a = αρc BS

Cb. (12)

It is shown in the Appendix that a = O(1).Because ρ is a function of x , in general equations (10) and (11) cannot be solved in

closed form. Previous studies have assumed that ρ is a constant, which is equivalentto ρ = 1. However, recent numerical work ([2], [3]) supports the hypothesis thatthe receptors occur in small regions of high density. This behavior occurs often inbiological systems (see Sect. 8.1 of [1]). Thus, let us consider the case of a “densitypulse” at a point x0. For x near x0, we let ρ(x) = O(ε−p) with p > 0. Because wehave defined

1∫

0

ρ(x) dx = 1, (13)

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the width of each pulse must be ε p. Thus

ρ = ε−pr(ξ), ξ = x − x0

ε p, and x0 = O(1).

For the regions outside of the pulse, we take the density to be o(1). It is most convenientto make the density transcendentally small. The pulses should be symmetric in ξ andnonnegative. Thus we define r as follows:

r(ξ) = h′′(ξ), h′′(ξ) transcendentally small as |ξ | → ∞, (14)

where we use the h′′ form for later algebraic convenience. Symmetry and positivitythen require that

h′′ ≥ 0, h′′(−ξ) = h′′(ξ).

Now we integrate up to obtain additional facts about h. First, upon integrating fromh′′ to h′, we obtain one arbitrary constant. Another constant may be chosen in orderto specify the amplitude of the pulse. Therefore, we choose

h′(−∞) = −1

2, h′(∞) = 1

2. (15)

If we integrate h′ up to obtain h, we obtain an arbitrary constant, which we set equalto zero by requiring that

h(ξ) = −ξ

2+ transcendentally small terms as ξ → −∞. (16)

At first blush, it may look as if this implies more than the selection of a constant.However, if h(ξ) ∼ −ξ/2 + O(ξ A), where A < 1, then taking the second derivativewould lead to a term in h′′ that decays algebraically, which would violate (14).

Lastly, we wish to point out some implications of our definition for h. First, wenote that the integral of an even function is an odd function plus a constant. However,since h′(−∞) = −h′(∞), we conclude that h′ is odd. Moreover, we know that theintegral of an odd function is even, so we have that

h(ξ) = ξ

2+ transcendentally small terms as ξ → ∞. (17)

Here are a few simple choices for the pulse. First, we can choose it to be a δ-function,in which case we have

h′′(ξ) = δ(ξ), h′(ξ) = H(ξ) − 1

2and h(ξ) = |ξ |

2, (18)

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where H(ξ) is the Heaviside step function. Alternatively, we could choose it as aGaussian, in which case we have

h′′(ξ) = 1√π

exp(−ξ2

),

h′(ξ) = erf(ξ)

2, and h(ξ) = 1

2

[ξerf(ξ) + 1√

πexp

(−ξ2

)]. (19)

We should point out that one can examine the δ-function postulate for ρ independentof the ε−p context, and it may be useful to do so in the future. The value of theperturbation approach is to be able to handle smooth pulses and provide correctionsto the δ-function case.

The following re-formulation of the derivative term in (12) will be used in the nextsection:

∂(F(C))

∂t= −F2(C)

∂

∂t

( ε

Cn

). (20)

3 Perturbation solution

In this section we use perturbation techniques to solve the problem defined by theEqs. (12), (10), (7), (8) and (9).

Solution of the cAMP diffusion problem

First we examine the effect of the pulses on C . (This will also establish a value for p.)Substituting (20) into (12), we obtain

∂C

∂t= ∂2C

∂x2 + aρF2(C)∂

∂t

( ε

Cn

). (21)

Motivated by the form of (21), in the outer region we let

C(x, t) = C0(x, t) + O(ε).

Substituting this into (21) and using our decay hypothesis about ρ, we have, to leadingorder,

∂C0

∂t= ∂2C0

∂x2 , (22)

which is not surprising. Biologically, we are stating that simple diffusion holds becausethere are no receptors in the outer region.

In the inner region we hypothesize the following perturbation expansion:

C(x, t) = c0(ξ, t) + εc1(ξ, t) + ε2c2(ξ, t) + o(ε2), (23)

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where the form is motivated by (21). Substituting this form and our definition for ρ

into (21), we have, to leading order,

∂c0

∂t= ε−2p ∂2(c0 + εc1 + ε2c2)

∂ξ2 + aε1−pr F2(c0)∂(c−n

0 )

∂t,

from which we conclude

∂2c0

∂ξ2 = 0, (24)

∂2c1

∂ξ2 = 0, (25)

and

∂c0

∂t= ε2(1−p) ∂

2c2

∂ξ2 + aε1−pr F2(c0)∂(c−n

0 )

∂t, (26)

where we have used the fact that F(C) = O(1). Therefore, the dominant balance isgiven by p = 1.

The matching conditions for c0 are

c0(−∞, t) = C0(x−0 , t), c0(∞, t) = C0(x+

0 , t).

Solving (24) subject to the above yields

c0(t) = C0(x0, t), (27)

and hence there is no layer in C0.In the regions where C = O(1) it is acceptable to use the following expansion

F(C) ∼ 1 − ε

Cn(28)

for C = O(1). The exact function and our approximation are shown in Fig. 1. Notethat the agreement is quite good for a wide range of C values. However, it is not goodfor C small which would correspond to small t . In many, but not all, cases, the effectsof the reaction term for t small are negligible. In Sect. 4 we develop a delay iterationto handle this difficulty.

If we now examine the spatial derivatives of the inner solution we have

∂

∂x(c0 + εc1 + · · · )(±∞, t) = ∂

∂x(C0 + · · · )(x±

0 , t).

Since ∂c1/∂x = (1/ε)∂c1/∂ξ and ∂c0/∂ξ = 0 from (27) we have

∂c1

∂ξ(−∞, t) = ∂C0

∂x(x−

0 , t) and∂c1

∂ξ(+∞, t) = ∂C0

∂x(x+

0 , t).

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0

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1

Fig. 1 Plot of F and a perturbation approximation

From (25) and the above

c1(ξ, t) = ξ∂C0

∂x(x0, t) + A(t) (29)

and hence there is no layer in ∂C0/∂x , either (Note that A(t) = C1(x0, t).). So nomeasurable layer occurs in the O(1) outer solution, subject to the small-t restrictions.

Thus we may simply solve (22) on the domain 0 ≤ x ≤ 1, subject to the boundaryand initial conditions (7) written in our new variable:

C0(x, 0) = 0, C0(0, t) = 1,∂C0

∂x(1, t) = 0, (30)

Biologically, we are saying that since the threshold concentration for reaction is sosmall (O(ε)) that once C = O(1) the reaction has already progressed to completionand it no longer affects the diffusion process.

There are two ways to solve our system: a Fourier series and a “Laplace series”,based on an expansion of a Laplace transform solution. The latter works best for smallt , which is where we will be mistrustful of our solution. Thus we derive the standardFourier series solution and obtain

C0(x, t) = 1 −∞∑j=0

4

(2 j + 1)πexp

(−

[(2 j + 1)π

2

]2

t

)sin

(x(2 j + 1)π

2

). (31)

Note that for t = O(1), our expression can be approximated well by only a few termsin the Fourier series.

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Solution of the membrane potential equations

Now we turn our attention to V . In the outer region, motivated by our previous choiceof p = 1, we let

V (x, t) = V0(x, t) + εV1(x, t) + o(ε). (32)

Substituting (32) into (10) and (8) and using our hypothesis about ρ, we have, toleading orders,

∂2V0

∂x2 = 0, V0(0, t) = −1,∂V0

∂x(1, t) = 0, (33)

and

∂2V1

∂x2 = 0, V1(0, t) = 0,∂V1

∂x(1, t) = 0. (34)

From the above we see that in the outer region we have a time-varying series of linearprofiles which matches the numerical solution in Fig. 2b of [2].

For the inner regions, we hypothesize the following perturbation expansion,

V (x, t) = v0(ξ, t) + εv1(ξ, t) + ε2v2(ξ, t) + o(ε2). (35)

Substituting our expansions into (10), we have, to leading three orders,

ε−2 ∂2(v0 + εv1 + ε2v2)

∂ξ2 = bε−1r F(c0 + εc1)(v0 + εv1 + ε2v2).

So,

∂2v0

∂ξ2 = 0, (36)

and, noting that c0 = C0,

∂2v1

∂ξ2 = bh′′F(C0)v0, (37)

as well as

∂2v2

∂ξ2 = bh′′[F(C0)v1 + c1 F ′(C0)v0]. (38)

Now we carefully consider the implications of our equations. In order to estimate theparameters in the inverse problem, we need to use all the time-dependent data we havebeen given. Thus, we needed to expand to O(ε2) in the above in order to get time

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dependence in our problem through C0. In particular, we see that nowhere does C0enter into our equations for the leading order of V . Without C0, there is no way fort to enter the problem. Hence we determine that V0 and v0 are independent of t . Thematching conditions for v0 are then given by

v0(−∞) = V0(x−0 ), v0(∞) = V0(x+

0 ).

Solving (36) subject to the above yields

v0 = V0(x), (39)

and hence there is no layer in V0. Rather, the layer is in the x-derivative of V , justas is seen in Fig. 2b of [2]. (Note that the value of V0 at the pulse is currently undeter-mined.)

Next we continue by examining v1. The matching conditions for v1 are given bymatching to our outer solution:

∂v1

∂ξ(−∞, t) = dV0

dx(x−

0 ),∂v1

∂ξ(∞, t) = dV0

dx(x+

0 ). (40)

Next we substitute V0 in for v0 in (37), then integrate once with a specific endpoint ofξ = −∞ to obtain

∂v1

∂ξ(ξ, t) − ∂v1

∂ξ(−∞, t) = bV0 F(C0)[h′(ξ) − h′(−∞)]

Now, from (15) and (40),

∂v1

∂ξ(ξ, t) − dV0

dx(x−

0 ) = bV0 F(C0)

[h′(ξ) + 1

2

]. (41)

By substituting ξ = ∞ into the above we obtain a jump condition on dV0/dx :

dV0

dx(x+

0 ) − dV0

dx(x−

0 ) = bV0 F(C0) (42)

using (40) again. Note that as far as the jump condition is concerned, the actual formof h does not come into play. Also note that (42) is exactly what we would obtain hadwe taken ρ to be a δ function without any consideration of ε.

Integrating (41) (but this time doing an indefinite integral), we have

v1(ξ, t) = ξdV0

dx(x−

0 ) + bV0 F(C0)

[h(ξ) + ξ

2

]+ A(t), (43)

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where A(t) is unknown for now. The matching conditions for v1 are more subtle.Expanding (32) near x0, we have

V (x0 ± εξ, t) = V0(x0) + εξdV0

dx(x±

0 ) + εV1(x±0 , t) + o(ε)

Since V (x0 + εξ, t) ∼= v0(ξ) + εv1(ξ, t) we find that for ξ → −∞ at the O(ε) level

v1(ξ, t) ∼= ξdV0

dx(x−

0 ) + V1(x−0 , t). (44)

Similarly, we have for ξ → +∞

v1(ξ, t) ∼= ξdV0

dx(x+

0 ) + V1(x+0 , t). (45)

So, letting ξ → −∞ and using (43) with (44), we have

ξdV0

dx(x−

0 ) + bV0 F(C0)

[h(−∞) + ξ

2

]+ A(t) − ξ

dV0

dx(x−

0 ) = V1(x−0 , t)

so

A(t) = V1(x−0 , t), (46)

where we have used (16). Letting ξ = +∞ in (43) and using (45), we have

dV0

dx(x−

0 )ξ + bV0 F(C0)

[h(∞) + ξ

2

]+ A(t) − ξ

dV0

dx(x+

0 ) = V1(x+0 , t).

Now, solving for dV0dx (x−

0 ) in (42) and substituting in the above,

ξdV0

dx(x+

0 ) − bV0 F(C0)ξ + bV0 F(C0)ξ + A(t) − ξdV0

dx(x+

0 ) = V1(x+0 , t)

where we used (17), so

A(t) = V1(x+0 , t). (47)

From (46) and (47), we see that V1 is continuous at x0, just as V0 is. Therefore thereis no boundary layer in V1. Rather, the layer is in the x-derivative of V1, just as in V0.In addition, our solution for v1 may be written as

v1(ξ, t) = ξdV0

dx(x−

0 ) + bV0 F(C0)

[h(ξ) + ξ

2

]+ V1, (48)

where as before we now drop the arguments on V1 since we know it is uniquely definedat x0 and independent of ξ .

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Substituting our formulas for v0 and v1, (39) and (48) into (38), we obtain

∂2v2

∂ξ2 = bh′′{

F(C0)

[ξ

dV0

dx(x−) + bV0 F(C0)(h(ξ) + ξ

2) + V1

]

+(

ξ∂C0

∂x+ C1

)F ′(C0)V0

}. (49)

As was done in (40) we find that

∂v2

∂ξ(∞, t) − ∂v2

∂ξ(−∞, t) = ∂V1

∂x(x+

0 , t) − ∂V1

∂x(x−

0 , t). (50)

Therefore, we may simply integrate (38) from ξ = −∞ to ξ = ∞ to obtain a jumpcondition for V1:

∂v2

∂ξ(∞, t)− ∂v2

∂ξ(−∞) = b

∞∫

−∞h′′ [F(C0)(bV0 F(C0)h(ξ)+V1)+F ′(C0)C1V0

]dξ,

where we have used the fact that h′′ is even. Continuing to simplify, we have from (50)

∂V1

∂x(x+

0 , t) − ∂V1

∂x(x−

0 , t)

= b2 F2(C0)V0γ + b(F(C0)V1 + F ′(C0)C1V0) with γ = 2

∞∫

0

h′′h dξ (51)

where we used the fact that∫ −∞∞ h′′ dξ = 1. Note that it is at this order that the actual

behavior of h, rather than just its behavior as |ξ | → ∞, becomes important.We now examine the parameter γ . For the two choices of h, the delta function and

Gaussian, we have

h′′ = δ(ξ) ⇒ γ = 0,

and

h′′ = 1√π

exp(−ξ2

)⇒ γ = 1√

2π.

We now obtain several helpful jump conditions for V ∼= V0 + εV1 by combining(42) and (51):

∂(V0 + εV1)

∂x(x+

0 , t) − ∂(V0 + εV1)

∂x(x−

0 , t) = b(V0 + εV1)F(C0) + εb2 F2(C0)V0γ

+ εF ′(C0)C1V0.

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We note from (28) that F ′(C0) = O(ε). Then dropping the O(ε2) terms, we have

∂V

∂x(x+

0 , t) − ∂V

∂x(x−

0 , t) = bV F(C0)(1 + εbF(C0)γ ). (52)

From (33) and (34) we have that V is piecewise linear. Using the BC’s we have

V (x0, t) + 1

x0= ∂V

∂xon 0 ≤ x < x0 and

∂V

∂x= 0 on x0 < x ≤ 1. (53)

Using the above in (52) and solving for V (x0, t), we obtain

V (x0, t) = 1

1 + bx0 F(C0)(1 + εbF(C0)γ ).

Now, from (9) and (53) we have

I = −∂V

∂x(0, t) = −∂V

∂x(x−

0 , t) = bF(C0)(1 + εbF(C0)γ )

1 + bx0 F(C0)(1 + εbF(C0)γ ), (54)

where we have used our expression above for V .

4 Numerical computations with the perturbation solution

We now use our perturbation solution approach to solve problems with simulated dataand with real data from S. J. Kleene’s lab (see [2] or [3] for more information on theexperiments). We examine the case where ρ ∼ δ and thus γ = 0. In this case we havefrom (54)

I (t) = bF(C0(x0, t))

1 + bF(C0(x0, t))x0. (55)

When considering situations with real data we note that the dimensional data current,I (t) tends to a limiting value as t → ∞, so it is natural to define

I∞ = limt→∞

[I (t)

Ic

].

Rewriting (55) with our form for F , we have

I (t) = b

1 + bx0 + εC−n0 (x0, t)

so it is natural to take I∞ = b(1+bx0)−1 in the perturbation expansion, which implies

b = I∞(1 − I∞x0)−1. The only unknown in I (t) in (55) now is the parameter x0.

We use current data to find a value for x0 that minimizes the sum of squares (over the

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available data points) of the differences between I (t) and the perturbation formula in(55).

Thus there are two parts of our numerical computation. First, C0 in (55) is computed.We used a sufficient number of terms in the separation of variables expansion (31) toensure an accurate representation of C0. The more major numerical computation inour procedure is the one-dimensional minimization. After determining an appropriatex0 = Lx0 we find a correction to b by noting that the experimental solutions tend toreach a steady state after t = 1. Therefore, we estimate I∞ with data at that time:

I∞ ∼= b

1 + bx0 + εC−n0 (x0, 1)

so b ∼= I∞1 − I∞x0

(1 + εC−n0 (x0, 1)).

Below we detail our results for a simulated example where we know the cilium’schannel distribution and an example with real data.

Example 1 (Simulated) Here we created current data by first solving a forward pro-blem with ρ(x) ∼ δ(x − x0) where x0 = 1.7 × 10−3 cm. We set ρ so it has 400 CNGchannels and the cilium length is L = 5.0 × 10−3 cm. The current resulting from thisforward problem was used as data for our perturbation solution procedure. We tookCb = 40 µM.

Our procedure found x0 = 1.76×10−3 cm and, from the b-value, the total numberof channels;

Total number of channels = b

Ra LgCNG P∼= 409 Channels.

Figure 2 has the true and perturbation currents which compare favorably in this case.

Example 2 Figure 3 displays the results from a case with real data. Here the pertur-bation procedure determined there were 290 channels and located the delta functiondistribution at x0 = 1.04 × 10−3 cm.

These two examples show results that are quite common. However, in some cases,especially for cilia with larger numbers of channels, the current residuals were not assmall.

Introducing a delay

As indicated above, the current from our perturbation approximation is not alwaysaccurate. This is primarily because at no time do we take into account the binding.Though it does occur only in a small region near the pulse for a small time, it does havean effect on the concentration (particularly the flux) for a time in the neighborhood ofthe pulse.

Handling the binding correctly would involve a complicated nonlinear boundarycondition on the flux. But our simulations show that in effect, the binding delays thediffusive process. In particular, we note that once enough time has elapsed, C reachesa value where the approximation in (27) holds and then the binding can be neglected.

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

Cur

rent

(pA

)

Time (s)

PerturbationRaw Data

Fig. 2 The 400 channel simulated case described in Example 1. Comparison of true current and approxi-mated current. Here Cb = 40µM and L = 5.0 × 10−3 cm

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

10

20

30

40

50

60

70

Cur

rent

(pA

)

Time (s)

PerturbationRaw Data

Fig. 3 Example 2 real data case. Here Cb = 10 µM and L = 5.0 × 10−3 cm. Comparison of true currentand approximated current

The correct nonlinear boundary condition on C would require numerical solution, asdiscussed above.

Thus, what we wish to do instead is introduce a delay into the concentration as usedin the computation of the current. In other words, since the current will depend on thetrue concentration at time t , it will depend on the computed concentration (which doesnot include the delay) at some time t − t , where t is the delay. Mathematically,we wish to replace (55) by

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I0 = − bF(C0(x, t − t))

1 + bF(C0(x, t − t))x0. (56)

To estimate the delay, we integrate (12) across the pulse:

0 =[∂C

∂x

]x+0

x−0

− a∂(F(C(x0, t)))

∂t, (57)

where we have used the definition of ρ.Next we examine the gradient terms. We assume that binding uses up all the flux

into the pulse during this short delay period, so the gradient at x+0 is zero. On the left,

computations of the concentration show an almost linear profile from C(0, t) = 1 toa very small value at x = x0. Thus we may estimate the gradient at x−

0 by −1/x0.Substituting these results into (57), we have

a∂(F(C(x0, t)))

∂t= 1

x0.

Then integrating this equation to our delay time t , we have

a[F(C(x0,t) − F(C(x0, 0))] = t

x0

so

t = aF(C(x0,t))x0, (58)

where we have used the initial condition for C .Thus the computation of the delay really calls for a choice of a threshold value of F .

Thus we wish to select a value of F corresponding to a C above (which is equivalentto a time past) which binding is a minimal effect. The first choice might be to pick thevalue of F corresponding to maximum binding. Since the binding term in (12) maybe written as

F ′(C)∂C

∂t,

(where the second term is assumed to be roughly the same size throughout the interval(0,t)), this corresponds to a maximum in F ′(C), or the value of F at which F ′′ = 0.

We hence calculate F ′′ in terms of F :

F ′′ = nF(1 − F)

C2 [n(1 − 2F) − 1] .

Therefore, the value of F at the inflection point of interest is given by

F = n − 1

2n.

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For the value of n we have chosen, this corresponds roughly to F ≈ 0.21. Then wewould substitute this value into (58) to obtain an estimate for t .

Simulations show that while this value of t does improve the fit, it is still toosmall. And upon reflection, we see that by choosing F at the inflection point, we areignoring only the first half of the binding (up to the maximum of F ′). Therefore, whatwe now wish to do is calculate the F corresponding to the latter inflection point of F ′.This would then allow us to ignore much more of the binding.

Calculating F (3), we have

F (3) = nF(1 − F)

C3

[6n2 F2 + 6n(1 − n)F + n2 − 3n + 2

].

Therefore, the latter value of F at which F (3) = 0 is given by

F = 3(n − 1) + √3n2 − 3

6n.

For the value of n we have chosen, this corresponds roughly to F ≈ 0.44. We foundthat the value 0.44 was too high and, therefore, used the value 1/3 which is intermediateto the earlier estimate of 0.21 and this new 0.44.

Now we can devise the following iterative scheme. Given a problem with no delay(t0 = 0), use our algorithm to compute estimates for b and x0. This can be done asin Examples 1 and 2. ρc (and hence a) may be calculated from the computed b value.Once a and x0 are calculated, a new t is calculated using (58):

tn+1 = F∗a(tn)x0(tn) ≡ F(tn), (59)

where we have chosen F∗ = 1/3. This iteration should then have a fixed point t∗which will give better estimates for b and x0 than t = 0. Unfortunately, we havefound that the iterative scheme in (59) has very slow convergence, with oscillationsabout the fixed point t∗. Thus we replace the scheme (59) with

tn+1 = (1 − ω)F(tn) + ωtn . (60)

(Note that fixed points of (60) are fixed points of (59).) We took ω = 1/2 to speedconvergence. We found that, typically, after 3–5 iterations the tn’s stabilized.

Below we have two examples where we have used this delay procedure. We evaluatea residual error for the current outputs for each case;

Residual =∑

i |IData(ti ) − IAppx(ti )|∑i |IData(ti )|

where ti are the data points.

Example 3 (Simulated with 1,600 channels) Figure 4 displays our results for datacreated from a delta function distribution located, as in Example 1, at 1.7 × 10−3 cm

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

50

100

150

Cur

rent

(pA

)

Time (s)

∆ t = 0 ∆ t = 5.76e−003Raw Data

Fig. 4 Example 3 simulated data case. Here Cb = 40 µM and L = 5.0 × 10−3 cm

and with 1,600 channels. The location and channel count found with delay t = 0 swas x0 = 2.1 × 10−3 cm with 4993 channels. However, 4 steps of the delay iterationprocedure we found t = 5.76 × 10−3 s and an accurate estimate of the channeldistribution with x0 = 1.66 × 10−3 cm with 1685 channels. Figure 4 reveals thesignificant improvement in the residual by using this delay procedure. The residualfor the t = 0 case was 0.137 while, after the iteration, with t = 5.76 × 10−3 was0.012; which is a tenfold improvement.

Example 4 (Real Data from the Kleene Lab) Figure 5 displays our results for realdata using the iteration scheme with 3 iterations. The cilium in this example was7×10−3 cm and we had Cb = 20 µM. Again the delay iteration produced an improvedapproximation with final t = 5.17×10−3 s. Here we had x0 = 1.70 ×10−3 cm and

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

20

40

60

80

100

120

140

160

Cur

rent

(pA

)

Time (s)

∆ t = 0 ∆ t = 5.17e−003Raw Data

Fig. 5 Example 4 real data case. Here Cb = 20 µM and L = 7.0 × 10−3 cm

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1385 channels. The residual for the t = 0 case was 0.099 while, after the iterations,the residual was 0.040 and the calculated delay was t = 5.17 × 10−3.

Acknowledgments D. A. French was partially supported during 2002-4 by an Interdisciplinary Grant inthe Mathematical Sciences (IGMS) from the National Science Foundation (NSF DMS 0207145) as well asthrough cost-sharing from the Taft Foundation, Department of Mathematical Sciences, Dean of Arts andSciences, and the Provost at the University of Cincinnati. Trips to the Mathematical Biosciences Institutewere beneficial to this work as is current support by NSF grant (DMS 0515989) with S. Kleene as Co-PI.D. A. Edwards was supported by NIGMS Grant 1R01GM067244-01. Portions of this research were com-pleted during a visit to the Mathematical Biosciences Institute.

Appendix

Now we use typical physical parameters to estimate our dimensionless parameters. Thereferences for the physical constants throughout are [2] and [3]. We choose Cb = 30µM from the following range of values used in experiments:

5 µM ≤ Cb ≤ 300 µM.

For K1/2 and n, we have

K1/2 = 1.7 µM and n = 1.7.

Thus, ε is

ε =(

K1/2

Cb

)n

= 7.60 × 10−3.

In order to calculate the characteristic current Ic, we need the following parameters:

Vb = 50 mV = 5 × 10−2 V, and Ra = 1.49 × 1011 �

cm.

In the calculation of Ra, we note that the intracellular resistance is given by Ra = Ri/A,where A is the cross-sectional area. The intracellular resistivity Ri is known to be 91.7�-cm. We took the diameter of the cilium to be 2.8×10−5 cm. We chose L = 3×10−3

cm from the following experimental range:

3 × 10−3 cm ≤ L ≤ 10 × 10−3 cm.

Then for the characteristic current we have

Ic = Vb

Ra L= 112 pA.

For the characteristic density, experimental values range between

1.4 × 105 channel

cm≤ ρc ≤ 106 channel

cm.

Again, we use a typical value, ρc = 3 × 105 channels/cm (900 channels).

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To calculate b, we also need the following values, given in the text: gCNG = 8.3 ×10−12(� · channel)−1 and P = 0.7. We thus obtain the following value for b:

b = Ra L2gCNG PVbρc = 2.35.

We may calculate the characteristic time scale once we have the following value:D = 2.7 × 10−6 cm2/s. Thus

tc = L2

D= 3.33 s.

Note that this value is somewhat long compared to the time scale evident in thenumerical simulations, but right on par with the physical experiments.

To calculate a, the only remaining parameter we need is α, which is given by

α = 1

NA A= 2.7 × 10−6 µM · cm

molecule

where NA = 6.0 × 1023 molecules/mole is Avogadro’s number. We also assignBS = 1.7 molecules/channel (number of binding sites needed to activate the CNGchannel current). With this we can now obtain the dimensionless parameter a,

a = αρc

CbBS = 4.6 × 10−2.

References

1. Fall, C.P., Marland, E.S., Wagner, J.M., Tyson, J.J.: Computational Cell Biology. Springer, New York(2000)

2. Flannery, R., French, D.A., Groetsch, C.W., Krantz, W.B., Kleene, S.J.: CNG channel distributions inthe cilia of frog olfactory receptor neurons. Math. Comp. Model. 43, 945–956 (2006)

3. Flannery, R., French, D.A., Kleene, S.J.: Clustering of CNG channels in grass frog olfactory cilia.Biophys. J. 91, 179–188 (2006)

4. Kleene, S.J.: Both external and internal calcium reduce the sensitivity of the cyclic-nucleotide-gatedchannels to cAMP. J. Neurophysiol. 81, 2675–2682 (1999)

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